Show that the normal at any point $\theta$ to the curve ` $x = a\cos \theta + a\theta \sin \theta ,y = a\sin \theta - a\theta \cos \theta$ is at a constant distance from the origin.

We have,
$x = a\cos \theta + a\theta \sin \theta$ …(i) and
$y = a\sin \theta - a\theta \cos \theta$ …(ii)

Differentiating (i) \& (ii) w.r.t. $\theta$, we get
$\cfrac{{dx}}{{d\theta }} = - a\sin \theta + a\{ \theta \cos \theta + \sin \theta \} = a\theta cos\theta$

and $\cfrac{{dy}}{{d\theta }} = a\cos \theta - a\{ \theta ( - \sin \theta ) + \cos \theta \} = a\theta \sin \theta$

Therefore the slope of the tangent at $\theta$ is given by
$\cfrac{{dy}}{{dx}} = \cfrac{{\cfrac{{dy}}{{d\theta }}}}{{\cfrac{{dx}}{{d\theta }}}} = \cfrac{{a\theta \sin \theta }}{{a\theta \cos \theta }} = \tan \theta$

$\Rightarrow$ Slope of the normal at $\theta$ is given by $- \cfrac{1}{{\cfrac{{dy}}{{d\theta }}}} = - \cfrac{1}{{\tan \theta }} = - \cot \theta$

The equation of the normal at $\theta$ is
$y - (a\sin \theta - a\theta \cos \theta ) = - \cfrac{{\cos \theta }}{{\sin \theta }}(x - (a\cos \theta + a\theta \sin \theta ))$

$\Rightarrow y\sin \theta - a{\sin ^2}\theta + a\theta \sin \theta \cos \theta$
$= - x\cos \theta + a{\cos ^2}\theta + a\theta \sin \theta \cos \theta$

$\Rightarrow x\cos \theta + y\sin \theta = a({\sin ^2}\theta + {\cos ^2}\theta )$
$\Rightarrow x\cos \theta + y\sin \theta = a$, which is in the normal form.

Its distance from the origin $= {\rm{ }}\cfrac{{| - a|}}{{\sqrt {{{\cos }^2}\theta + si{n^2}\theta } }} = \cfrac{a}{{\sqrt 1 }} = a$, which is constant.